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Number 105
Calculations on the Photoelectric Effect
When a metal surface is illuminated with light, electrons can be emitted from the metal surface, provided the frequency of the light is
above a certain value. This is called the photoelectric effect.
The speeds (and energies) of the electrons will depend upon how
much of their energy they used getting to the surface. The fastest
electrons will be the ones that are already at the surface, so that B is
zero. This gives us our final formula, relating only to the fastest
electrons, and the one we will use:
The photoelectric effect is one of the key pieces of
evidence that light can behave as a stream of particles called
photons, rather than as a wave. If light were behaving as a
wave, then the emission of electrons would behave very
differently. This is beyond the scope of this factsheet.
Fig 1
hf = W + KEmax
or, hf = W + ½mv2max
energy
delivered by
photon (A)
Exam Hint: when calculating speeds from kinetic energies as
we will do later, the KE must be in Joules. However, as you have
already seen, the work function and photon energies are often
given in electronvolts (eV). It is therefore vital that you:
• remember at all times which units you are using
• are able to convert between them: 1 eV = 1.6 × 10-19 J
remaining energy
ends up as kinetic
energy of electron
(D)
Energy given to electron:
some is used to get electron to surface (B)
and some is used by electron in escaping surface (C)
Worked example
A metal surface with a work function of 4.0 electronvolts is
illuminated by light of wavelength 200 nm. What is the maximum
velocity of the photoelectrons produced?
speed of light
= 3 × 108 m s-1
Planck’s constant = 6.6 × 10-34 J s
mass of electron = 9.1 × 10-31 kg
As you can see in Fig 1, behind the photoelectric effect is a really
simple idea:
The first step is to calculate the frequency of the light from its
wavelength
v = fλ, so
v
3 × 108 m s-1
f=
=
λ
200 × 10-9 m
(don’t forget the powers of ten when
= 1.5 × 1015 Hz
dealing with nano, micro, milli etc)
Energy delivered by the photon = energy delivered to the electron
(A in the diagram)
Of the energy delivered to the electron:
• some is used in getting electron to the metal surface (B in the
diagram)
• some is used in getting electron to escape the metal surface (C)
• whatever energy remains appears as kinetic energy of the
electron (D)
So, very simply:
Now we can use either of the formulas above.
The first (hf = W + KEmax) will work but requires a two step process
– you will have to use KE = ½ m v2 afterwards to convert KE into
speed, so we will use the second formula.
A=B+C+D
hf = W + ½mv2max
A : is the energy of the photon. This is proportional to the frequency
of the light, such that Energy of photon = hf
where h is Planck’s constant = 6.63 × 10-34 Js
(6.6×10-34 J s) × (1.5×1015 s-1) = (4.0 × 1.6×10-19 J) + (0.5 × 9.1×10-31 kg × v2)
(Notice that I changed the work function into Joules from
electronvolts)
C: is the energy needed for an electron to escape the surface. This
is known as the work function, and is a constant for a given
metal. For example, the work function of zinc is 4.24 eV, and that
of caesium is 1.35 eV. The symbol for work function W or φ .
giving v = 8.8 × 105 m s-1
Exam Hint: Check you can get this result with your calculator.
It is very common for students to do all the hard work and then
forget to square-root v2 at the end. Don’t fall into this trap!
D is the kinetic energy of the electron and is thus given by:
KE = ½ m v2
So: Energy delivered by the photon = energy delivered to the
electron now turns into:
energy to get
hf =
+ W + ½ m v2
to surface, B
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Physics Factsheet
105. Calculations on the Photoelectric Effect
Graphs in the photoelectric effect
The equation of a straight line is:
y=m x+c
The formula hf = W + KEmax results in the following graph
maximum energy of emitted electrons (eV)
Fig 2
The photoelectric equation can be written KE = h f - W
Comparing the two, and looking at Fig 3 shows that:
• the gradient of the line gives a value for Planck’s constant
• the threshold frequency, f0, can be read off the x-axis
• the work function can be obtained by continuing the line and
reading off the negative y-intercept. Notice that the work
function can also be obtained from the threshold frequency:
W = h f0
8
7
6
5
4
A photoelectron is not a different particle from an electron.
It is just an electron that has been emitted from a metal by a
photon. It is still a normal electron!
3
2
1
P
0
0
1
Q
R
Stopping potential
S
If a negative electrode is placed near the emitting metal surface,
then it will repel the photoelectrons. If it is negative enough, then
the photoelectrons will not reach it.
2 3 4
5 6 7
8 9
energy of incident photon (eV)
metal electrode
from which
photoelectrons
are released
Consider 4 incident photons P, Q, R and S, which have different
energies (and frequencies)
• Photon P does not have enough energy to cause any
photoemission of electrons.
• Photon Q has just enough energy to cause photoemission. There
is no ‘spare energy’ left over, so the photoelectrons have no KE
– all the energy went into leaving the surface. This is of course
the work function, so the work function of this particular metal
is 4 eV.
• Photon R had 6 eV. 4 eV of those were used in getting from the
surface, so the maximum KE possible is 2 eV.
• Photon S had 9 eV. 4 eV of those were used in getting from the
surface, so the maximum KE possible is 5 eV
• By considering R and S, you should be able to see that the
gradient of the graph is 1, or ‘45 degrees’.
collecting electrode
photoelectron
vacuum
V
galvanometer
(sensitive ammeter)
variable voltage supply
The circuit above shows a photocell. As the voltage of the supply
is increased, more and more electrons are repelled back, and the
galvanometer reading falls. Only the fastest electrons reach the
collecting electrode. When the supply voltage reaches a value known
as the stopping potential, even the fastest electrons can not reach
the collecting electrode and the current falls to zero.
Note that if the energy of the photon is not great enough
for an electron to escape the surface, then no electrons will be
emitted. This is why there is a threshold frequency: above that
frequency electrons are emitted, below it they are not.
In energy terms, the electrical potential energy of an electron near
the collecting electrode has risen higher than the kinetic energy of
the photoelectrons, which therefore cannot reach the electrode.
(This is the electrical equivalent of a very simple gravitational idea:
if you place a target above you on a wall you may be able to hit it by
throwing a ball, but if you raise the target there will come point
where you can not give the ball enough KE to reach.)
Sometimes this graph is plotted with the frequency of the light on
the x-axis rather than the photon energy. In this case the gradient of
the line is not 1.
Fig 3
maximum energy of emitted electrons (J)
incident
light
The electrical energy transferred to an electron crossing from one
electrode to the other is qV, where q is the charge on an electron and
V is the potential difference between the electrodes. At the stopping
potential Vstop, the fastest electrons are just prevented from making
it, so:
threshold
frequency
gradient of line is
equal to Planck's
constant
qVstop = ½mv2max
which equals hf - W
frequency of incident light (Hz)
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Physics Factsheet
105. Calculations on the Photoelectric Effect
Practice Questions
Answers
For the following questions, you may use the data below:
speed of light
= 3 × 108 m s-1
mass of electron
= 9.1 × 10-31 kg
charge on electron = 1.6 × 10-19 C
Planck’s constant = 6.6 × 10-34 Js
1. (a)
(b)
(c)
(d)
1. Light of frequency 9.2 × 1014 Hz falls upon a metal with a work
function of 2.5 eV.
(a) Calculate the maximum kinetic energy of the resulting
photoelectrons
(b) Calculate the maximum speed of the resulting photoelectrons
(c) What is the threshold frequency for photoelectric emission?
(d) If a nearby electrode is made negative using a potential
difference V, what value of V is required to stop any
photoelectrons reaching it?
2.1 × 10-19 J
6.7 × 105 m s-1
6.1 × 1014 Hz
1.3 V
2. 2.9 × 10-7 m (295 nm)
3. (a) 1.3 × 10-19 J (0.81 eV)
(b) must be greater than 1.5 × 10-6 m
4. (a) approx 0.7 eV
(b) approx 6.7 x 10-34 Js
5. 6.6 × 10-19 J (4.1 eV)
2. The work function of zinc is 4.2 eV. What is the maximum
wavelength of light that will cause photoemission of electrons
from zinc?
6. 1.9 V
7. 0.19 V
3. Light of wavelength 0.60 µm incident upon a metal surface ejects
photoelectrons with kinetic energies up to a maximum value of
2.0 × 10-19 J.
(a) What is the work function of the metal?
(b) If a beam of light causes no photoelectrons to be emitted,
what can you say about its wavelength?
4. The graph below shows how the maximum KE of photoelectrons
depends on the frequency of incident radiation for a certain
metal.
KEmax/eV
3
2
1
0
0
1
2 3 4
5
6 7
f/1014 Hz
(a) Use the graph to estimate the work function of the metal
(b) Use the graph to obtain a value for Planck’s constant (harder)
5. A sample of magnesium is used as an electrode in a photocell. It
is illuminated with u-v light of wavelength 225 nm, and a current
flows in the photocell. The current can be reduced to zero by
making the other electrode negative using a potential difference
of 1.4 V. Calculate the work function of magnesium.
6. The work function of caesium is 1.35 eV. A photocell contains a
caesium surface that is illuminated with light of wavelength 380
nm. What potential difference must be applied the the cell to
just prevent a current passing through it?
Acknowledgements:
This Physics Factsheet was researched and written by Michael Lingard
The Curriculum Press,Bank House, 105 King Street,Wellington, Shropshire,
TF1 1NU
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ISSN 1351-5136
7. A metal surface with a work function of 2.9 eV is illuminated with
light of wavelength 400 nm. What will be the stopping potential
for the photoelectrons?
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